\subsection*{2012-06-21}
The question we want to answer is: what is the most general form of a metric that is both time-reversal and time-translation invariant, and in addition its spatial slices (const $z$ \& $t$)  are $\mathbb{R}^n$ or $S^n$. 
The space in question is $d+2$ dimensional, and we call the coordinates $(t, x_i, z)$.

Time-translation means we can choose a t coord st all the coefficients are t-independent. Let's write down a most general metric
\begin{equation}
ds^2 = f(t,x,z)\; dt^2 + v_i(t,x,z)\; dt\, dx^i + 2 u(t,x,z) \; dt\, dz + h_{ij}(t,x,z) \; dx^i\, dx^j + 2 w_i(t,x,z)\; dx^i\, dz + q(t,x,z)\; dz^2
\end{equation}

We wish write the metric in coordinates that will make the symmetries of the metric manifest. 
Imposing time-reversal immediately kills the $dx\,dt$ and $dz\,dt$ terms, as $dt \rightarrow -dt$ is not time invariant. 

Next, time-translation invariance guarantees the existance of a timelike vector field $X$ that generates an isometry. 
That is, $X$ is a Killing vector field and the metric satisfies $\mathcal{L}_X g = 0$.
The isometry allows us to assert that all metric coefficients be independent of $t$: a time coordinate that is an explicit realization of this is the parameter along the integral curves of $X$.

Moreover, since we want $SO(d+1)$ symmetry on the spacial slices, the $h_{ij}$ factorizes as $h_{ij}(x)\,d\Omega^2\, H(z)$. 
Furthermore, since the $dx\,dz$ term breaks the rotation symmetry, so we set it to zero. 
By a similar argument, the remaining $dt^2$ and $dz^2$ terms must be function of $z$ only. 

Notice that in the calculations did so far (2+1 asymptotically AdS space), we did not impose the $SO(d+1)$ symmetry by hand (i.e. we let the $dt^2$ be dependent on $x$), and the Einstein equations killed that $x$ dependence automtically. 
This is for different reasons, and Allan said it is subtle and has to do with there being only one $x$ dimension. Boh.

The resulting form of the metric is:
\begin{equation}
ds^2 = f(z)\, dt^2 + h(z)\, h_ij(x)\, dx^i\, dx^j + q(z)\,dz^2
\end{equation}
 
What about residual gauge symmetries? 
The most general coordinate transformation we can make is: $t' = f_t(t, x, z)$, $x' = f_x(t,x,z)$, $z' = f_z(t,x,z)$. 
However, any transformation that mixes up the time coordinate will destroy the manifest time symmetries. 
Similarly, transformation mixing $x$ are out of the question as well.
Changing the $z$ coordinate is all right though: $z' = z'(z)$.  
Locally, we can make the $z$ coordinate whatever we please (so long as the coordinate transformations are not singular), and still retain the form of the metric. 
Whether the resulting coordinate extends to the whole manifold is a separate issue, and we mau have to work-patch-wise on our manifold. 
Most generally, we should remain agnostic with regard to the precise choice of $g_{zz}$, as we might not know what the most convenient choice is, and compute the solution of EFE with this in mind, and set some coefficient to some convenient function at our discretion.

When we did this calculation in 2+1 dimensions, we set $q(z)$ to $1/z^2$, implicitly making our space asymptotically AdS (i.e. constant negative curvature). 
We then solved EFE order by order and got the metric be:
\begin{equation}
ds^2 = \tfrac{1}{z^2} \left[\left( f(0) - \tfrac{a f(0) z^2}{2h(0)} + \tfrac{f(0) a^2 z^4}{16h(0)^2}\right)\, dt^2 + \left(h(0) + \tfrac{1}{2} \tfrac{a z^2 + a^2 z^4}{16h(0)}\right) dx^2 + dz^2  \right]
\end{equation}

where $f(0)$ and $h(0)$ are the boundary values of the metric, and $a$ is an arbitrary degree of freedom with dimensions mass squared.

We then evaluated the on-shell action (Einstein-Hilbert) and got three divergences:
\begin{enumerate}
\item We integrated $dx$ to get an IR divergence (since we assumed $x$ is not a circle). 
This is not a problem though -- our action is extensive, so we should maybe be more interested in some sort of action density.
\item We integrated $a\, z\, dz$ to get an (a $z^2$) term which gives an IR divergence from the $z=\infty$. 
Interestingly, this divergence is controlled by our coefficient $a$. 
Note that if $a=0$, we simply have AdS everywhere. 
Although in 2+1 dimensions, our Ricci scalar was a constant, as Allan mentioned in 2+1 dimensions the mass manifests itself with a point singularity (e.g. a cone if our c.c. is 0) so that the space is still {\emph{locally}} AdS (or flat or whatever), but we have some mass singularity. 
So the fact that $R$ is constant does not mean we do not have a black hole horizon in 2+1 dimensions. 
The interpretation of this divergence goes something like this -- we integrated to z=infinity, which is a singularity from a BH. 
In the boundary theory, this will have a different interpretation which we will talk about Monday

\item We integrated $1/z^3 dz$ to get a $(1/z^2)$ UV divergence at $z=0$. 
This term can be killed by adding a counter-term -- the Gibbons-Hawking term. 
We added this formulaically from Wikipedia and indeed the $1/z^2$ divergence was killed. The GH term is something we add to the boundary of our manifold that looks very similar in structure to EH action. Maybe we can add a cc? What's up with this $K$ thing?
\end{enumerate}

 

{\bf{Homework}} for Monday is:
\begin{enumerate}
\item Find the metric with SO(d+1) isometry in 4+1, 5+1 dimensions, and compute the Ricci tensor and scalar (and generalize?). We should get terms with $d-2$ coefficients, which explain why our $R$ was constant in 2+1 dimensions. $R$ is not constant in >3 dimensions.
\item Why must we have boundary terms in GR? Why is the GH term part of the answer? What sets the coefficient of GH? What about the boundary cc? More generally what can appear in a boundary Lagrangian? (basically -- what's up with this gibbons hawking term?)
\item In 2+1, set a=0 and compute the on-shell action when $ds^2 = -dt^2 f(x,z) + 2 dt dx v(x,z) + h(x,z) dx^2 + dz^2/z^2$ st the space is asymptotically AdS. In other words, we relax the time-reversal and SO(d+1) isometry conditions (e.g. physical interpretation: turn on a magnetic field at time t)
\item Next time we will also talk about what problem we will solve in the next months.
\end{enumerate}

